Metamath Proof Explorer


Theorem sinaover2ne0

Description: If A in ( 0 , 2 _pi ) then sin ( A / 2 ) is not 0 . (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Assertion sinaover2ne0 A 0 2 π sin A 2 0

Proof

Step Hyp Ref Expression
1 elioore A 0 2 π A
2 1 recnd A 0 2 π A
3 2cnd A 0 2 π 2
4 picn π
5 4 a1i A 0 2 π π
6 2ne0 2 0
7 6 a1i A 0 2 π 2 0
8 pire π
9 pipos 0 < π
10 8 9 gt0ne0ii π 0
11 10 a1i A 0 2 π π 0
12 2 3 5 7 11 divdiv1d A 0 2 π A 2 π = A 2 π
13 0zd A 0 2 π 0
14 2re 2
15 14 8 remulcli 2 π
16 15 a1i A 0 2 π 2 π
17 0xr 0 *
18 17 a1i A 0 2 π 0 *
19 16 rexrd A 0 2 π 2 π *
20 id A 0 2 π A 0 2 π
21 ioogtlb 0 * 2 π * A 0 2 π 0 < A
22 18 19 20 21 syl3anc A 0 2 π 0 < A
23 2pos 0 < 2
24 14 8 23 9 mulgt0ii 0 < 2 π
25 24 a1i A 0 2 π 0 < 2 π
26 1 16 22 25 divgt0d A 0 2 π 0 < A 2 π
27 1rp 1 +
28 27 a1i A 0 2 π 1 +
29 16 25 elrpd A 0 2 π 2 π +
30 2 div1d A 0 2 π A 1 = A
31 iooltub 0 * 2 π * A 0 2 π A < 2 π
32 18 19 20 31 syl3anc A 0 2 π A < 2 π
33 30 32 eqbrtrd A 0 2 π A 1 < 2 π
34 1 28 29 33 ltdiv23d A 0 2 π A 2 π < 1
35 1e0p1 1 = 0 + 1
36 34 35 breqtrdi A 0 2 π A 2 π < 0 + 1
37 btwnnz 0 0 < A 2 π A 2 π < 0 + 1 ¬ A 2 π
38 13 26 36 37 syl3anc A 0 2 π ¬ A 2 π
39 12 38 eqneltrd A 0 2 π ¬ A 2 π
40 2 halfcld A 0 2 π A 2
41 sineq0 A 2 sin A 2 = 0 A 2 π
42 40 41 syl A 0 2 π sin A 2 = 0 A 2 π
43 39 42 mtbird A 0 2 π ¬ sin A 2 = 0
44 43 neqned A 0 2 π sin A 2 0